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Chapter 11: Problem 3

Why can we not guarantee that BCNF relation schemas will be produced by dependency-preserving decompositions of non-BCNF relation schemas? Give a counterexample to illustrate this point.

Short Answer

Expert verified
BCNF and dependency-preservation can contradict each other in the context of decomposition of a non-BCNF schema. This is due to the strict condition of BCNF where 'every determiner must be a candidate key', which does not always guarantee the maintenance of all functional dependencies post-decomposition. A counterexample is a schema R with attributes {A, B, C} and functional dependencies {A -> B, B -> C}, which upon BCNF decomposition into R1 {A, B} and R2 {B, C} loses the dependency {A -> C} of the original schema.

Step by step solution

01

Understanding BCNF

BCNF, or Boyce-Codd Normal Form, is a standard for the design of a relational database schema that avoids certain anomalies when inserting, deleting, and updating data. A table is in BCNF if it is in 3NF (Third Normal Form) and for every non-trivial functional dependency \(X \rightarrow Y, X\) is a superkey. Superkey basically means a set of attributes that uniquely identifies tuples in a relation.
02

Understanding Dependency-Preserving decomposition

Dependency-preserving decomposition ensures minimizing redundancy and chances of anomalies while maintaining all of the functional dependencies after normalization, which is really important in relational databases. This lets one enforce constraints (functional dependencies) of the original schema in the decomposed schema.
03

Identifying potential conflict

The conflict between BCNF and dependency-preservation comes into play when performing decompositions on a non-BCNF schema. BCNF may not be able to maintain dependency preservation upon decomposing non-BCNF schemas because of its strict condition of 'every determiner being a candidate key'. This is because BCNF does not necessarily take into account the need to maintain all functional dependencies at the time of decomposition.
04

Providing a counterexample

Let's take a schema R with attributes \( \{A, B, C\} \) and functional dependencies \( \{A \rightarrow B, B \rightarrow C\} \). This schema is not in BCNF since \(B\), which is not a superkey, determines \(C\). However, if we decompose it into BCNF, we might get two schemas: \(R1\{A, B\}\) with \({A \rightarrow B}\) and \(R2\{B, C\}\) with \({B \rightarrow C}\). We can observe that the dependency \({A \rightarrow C}\) from the original schema is lost in the decomposed one. Hence, this is a counterexample where a BCNF decomposition is not dependency-preserving.

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Most popular questions from this chapter

What is meant by the attribute preservation condition on a decomposition?

What is a multivalued dependency? What type of constraint does it specify? When does it arise?

What is the lossless (or nonadditive) join property of a decomposition? Why is it important?

, YEAR, PRICE, MANUF_PLANT, COLOR), which is abbreviated as REFRIG(M, Y, P, MP, C), and the following set \(F\) of functional dependenci鈥� # Consider the relation REFRIC(MODEL#, YEAR, PRICE, MANUF_PLANT, COLOR), which is abbreviated as REFRIG(M, Y, P, MP, C), and the following set \(F\) of functional dependencies: \(F=\\{M \rightarrow M P,\\{M, Y\\} \rightarrow P, M P \rightarrow C\\}\) a. Evaluate each of the following as a candidate key for REFRIG, giving reasons why it can or cannot be a key: \(\\{\mathrm{M}\\},\\{\mathrm{M}, \mathrm{Y}\\},\\{\mathrm{M}, \mathrm{C}\\}\) b. Based on the above key determination, state whether the relation REFRIG is in \(3 \mathrm{NF}\) and in BCNF, giving proper reasons. c. Consider the decomposition of REFRIG into \(D=\\{\mathrm{R} 1(\mathrm{M}, \mathrm{Y}, \mathrm{P}), \mathrm{R} 2(\mathrm{M}, \mathrm{MP}, \mathrm{C})\\}\) Is this decomposition lossless? Show why. (You may consult the test under Property LJ 1 in Section \(11.1 .4 .\)

What types of constraints are inclusion dependencies meant to represent?
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